If x is not equal to zero , is 1/x > 1 ?
(1) y/x > y .
(2) x^3 > x^2 .
source : Kaplan.
1/x > 1 ?
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I'll give this a shot. In order for 1/x to be > 1, x must be 0<x<1. So we can rephrase the stem by saying is 0<x<1?
A) Insufficient because we know nothing about y
if Y = 2, x could be 1
if Y = -2, x could be -1
B) Sufficient
In order for x^3 to be > x^2, X must be > 1
for example
if x = -1, (-1)^3 = -1 which is not greater than (-1)^2 =1
if x = 1/2, (1/2)^3 = 1/8 which is not greater than (1/2)^2 = 1/4
So x must be >1
So B
A) Insufficient because we know nothing about y
if Y = 2, x could be 1
if Y = -2, x could be -1
B) Sufficient
In order for x^3 to be > x^2, X must be > 1
for example
if x = -1, (-1)^3 = -1 which is not greater than (-1)^2 =1
if x = 1/2, (1/2)^3 = 1/8 which is not greater than (1/2)^2 = 1/4
So x must be >1
So B
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we need to determine if the statement is sufficient to answer
1/x > 1?
(1) y/x > y .
=>1/x >1
so (1) is sufficient
(2) x^3 > x^2 .
x^3 - x^2 >0
=>x^2 (x-1) > 0
=>x^2 > 0 or x-1 > 0
so (2) is insufficient
I think A should be the answer
1/x > 1?
(1) y/x > y .
=>1/x >1
so (1) is sufficient
(2) x^3 > x^2 .
x^3 - x^2 >0
=>x^2 (x-1) > 0
=>x^2 > 0 or x-1 > 0
so (2) is insufficient
I think A should be the answer
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I don't believe that this is accurate. We don't know the sign of y so we can't divide y on both sides of the equation. If y were negative we would have to flip the sign.
I would be interested in seeing the OA
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IMO B is right here
St1 Here the value of x depends upon the value of y. Let's see three scenarios below.
a. y = 0; x can be anything other than 0
b. y is -ve; x can be +ve or -ve (y= -2; x = 2; -1 > -2. OR y = -3; x = -3; 1>-3)
c. y is +ve; x must be greater than 0 and less than 1 (y=2; x=0.5; 4>2)
St 2 x^3 > x^2. If x is -ve then x^3 must be less than x^2 so given condition will be wrong. It is not correct. if x is greater than 0 and less than 1 then x^3 will be less than x^2. If x = 1 then x^3 will be equal to x^2. In all these three scenarios given sentence fails. Therefore, to satisfy given condition or statement x must be greater than 1, and this is the correct answer
What is OA?
St1 Here the value of x depends upon the value of y. Let's see three scenarios below.
a. y = 0; x can be anything other than 0
b. y is -ve; x can be +ve or -ve (y= -2; x = 2; -1 > -2. OR y = -3; x = -3; 1>-3)
c. y is +ve; x must be greater than 0 and less than 1 (y=2; x=0.5; 4>2)
St 2 x^3 > x^2. If x is -ve then x^3 must be less than x^2 so given condition will be wrong. It is not correct. if x is greater than 0 and less than 1 then x^3 will be less than x^2. If x = 1 then x^3 will be equal to x^2. In all these three scenarios given sentence fails. Therefore, to satisfy given condition or statement x must be greater than 1, and this is the correct answer
What is OA?
rockeyb wrote:If x is not equal to zero , is 1/x > 1 ?
(1) y/x > y .
(2) x^3 > x^2 .
source : Kaplan.
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- rockeyb
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Peter welcome to this forum . I am sorry I should have typed in the complete positive instead of just +ve . It never occurred to me that some one might not realize what +ve stands for ? Sorry for that once again .PeterFaulkner wrote:btw I am new here...what does "ve" mean?
Your explanation above is perfect and [spoiler]OA = B.[/spoiler]
"Know thyself" and "Nothing in excess"
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Thanks for the warm welcome!rockeyb wrote:Peter welcome to this forum . I am sorry I should have typed in the complete positive instead of just +ve . It never occurred to me that some one might not realize what +ve stands for ? Sorry for that once again .PeterFaulkner wrote:btw I am new here...what does "ve" mean?
Your explanation above is perfect and [spoiler]OA = B.[/spoiler]
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This is how i got started :
1) y\x>y
If you divide both sides by 1\y =>1\x>1 , wont that be sufficient as well?? - correct me if i am wrong
2)also sufficient
So the answer would be D??[/list][/i]
1) y\x>y
If you divide both sides by 1\y =>1\x>1 , wont that be sufficient as well?? - correct me if i am wrong
2)also sufficient
So the answer would be D??[/list][/i]
- ajith
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If You divide the inequality by a negative number the direction of the inequality Changesnisha.menon294 wrote:This is how i got started :
1) y\x>y
If you divide both sides by 1\y =>1\x>1 , wont that be sufficient as well?? - correct me if i am wrong
For example, if -5x > -5y then x<y (direction of the inequality changes)
In this case you do not y is negative or positive so, we can't be sure about the sign
So A is not sufficient
Last edited by ajith on Tue Apr 13, 2010 3:32 am, edited 1 time in total.
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Fell for the trap. Never multiply or divide both sides on an inequality by an unknown. x and y are unkown till we solve the eqs.nisha.menon294 wrote:This is how i got started :
1) y\x>y
If you divide both sides by 1\y =>1\x>1 , wont that be sufficient as well?? - correct me if i am wrong2)also sufficient
So the answer would be D??[/list][/i]
If you have -2x < 6 y can divide by -2 , but there is more to it.
PeterFaulkner post is enough to rule out 1.)
@ [email protected] x^3 > x^2 x^3 - x^2 >0
=>x^2 (x-1) > 0 =>x^2 > 0 or x-1 > 0 so (2) is insufficient
why x² is always +ve then (x-1) has to be +ve otherwise x^2 (x-1) < 0
so x-1 > 0 or x > 1 so 1/x < 1 Sufficient
I like you method but interpreted it differently. Thanks to all for helping me solve this.